LeetCode - Letter Combinations of a Phone Number

Letter Combinations of a Phone Number

2013.12.6 19:52

Given a digit string, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below.

Input:Digit string "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.

Solution:

　　Given a certain digit string and the corresponding mapping between digits [0-9] and letters [a-z],  return all the possible letter combinatios that map to this digit string. Since all the combinations must be enumerated, using DFS is efficient in coding. A mapping array between  digits and letters is needed.

　　Code segment is quite short and need no more explanation.

　　Time complexity is BIG_PI(i in range(0, len(digit_str))){number of mapping for digit_str[i]}, space complexity is O(len(digit_str)).

Accepted code:

// 7CE, 2RE, 1WA, 1AC, too bad~
#include <string>
using namespace std;

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        result.clear();
        
        dfs(digits, 0, "");
        
        return result;
    }
private:
    vector<string> result;
    string dict[10] = {
        " ", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"
    };
    
    void dfs(string &digits, int idx, string str) {
        if(idx == digits.length()){
            result.push_back(str);
            // 1RE here, didn't return
            return;
        }
        
        // 7CE here, dict[digits[idx] - '0'], type mismatch
        for(int i = 0; i < dict[digits[idx] - '0'].length(); ++i){
            // 1RE here, 1WA here, wrong index
            dfs(digits, idx + 1, str + dict[digits[idx] - '0'][i]);
        }
    }
};